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Chemistry

Elevation of Boiling Point

Gk Scientist June 6, 2022 No Comments
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Elevation of Boiling Point:

The temperature of a substance (liquid) at which the vapour pressure of a liquid is equal to atmospheric pressure is called the boiling point of the liquid. It is characteristic of each liquid. For example- the boiling point of H2O is 373 K.

We know that the vapour pressure of a solution containing a non-volatile solute is lower than the vapour pressure of the pure solvent. Thus, such solutions have to be heated more to boil the solution.

Elevation of Boiling Point - Elevation of Boiling Point

This increase in temperature or the additional temperature required to boil such solution is called elevation in boiling point. If T°b is the boiling point of the solvent and Tb is the boiling point of the solution, then, ΔTb is the elevation in boiling point which can be mathematically written as-

ΔTb = (Tb – T°b)

This elevation in boiling point is colligative property as the elevation in boiling point is directly proportional to the lowering of vapour pressure (ΔP),

i.e., ΔTb ∝ ΔP ……….(i)

But according to Raoult’s Law,
ΔP ∝ xB ……….(ii) [Where xb is the mole fraction of the solute]

From (i) and (ii), ΔTb ∝ xB
⇒ ΔTb = k . xB [Where k is the constant of proportionality]
⇒ ΔTb = (k . nB) / (nA + nB) = k (nB/nA)
Since solution is dilute and nB << nA [as ΔTb is colligative property]
⇒ ΔTb = k . nB / (WA/MA) = k . MAnB / WA ……….(iii)
[Where WA = Weight of solvent and MA = Molecular Mass of solvent]

If the weight of solvent WA = 1 kg; then nB / WA = molality (m) of the solution.

Therefore, ΔTb = k . MA . m = kb . m ……….(iv)
[Where kb = k . MA is another constant]

This new constant kb is called molal elevation constant or ebullioscopic constant.

If m = 1 ⇒ ΔTb = kb

Hence, the molal elevation constant is the elevation in boiling point when the molality of the solution is unity. This constant has a characteristic value for each liquid which changes with temperature. Its unit is K/m or K kg mol-1.

Example- For H2O, kb = 0.52 K kg mol-1 at 373 K.

Calculation of Molecular Mass of Solute from ΔTb:

From relation (iv), ΔTb = kb . m = kb (nB / WA) ……….(v)
Since molality is the number of moles of solute in 1000 gms of solvent,
Therefore, m = (WB/MB) X (1000/WA) ……….(vi)
Using relation (vi) in (v), we get
ΔTb = (kb . WB . 1000) / (MB . WA)
Therefore, MB = (kb . WB . 1000) / (ΔTb . WA)
Hence molar masses of non-volatile solute are determined.

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